Assume that $H$ is a Lebesgue measurable additive subgroup of $\mathbb{R}$. Is $H$ necessarily a Borel subset of $\mathbb{R}$?
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4$\begingroup$ If $H$ has positive measure, then $H-H=H$, thus 0 is interior point of $H$ (Steinhaus theorem) and $H=\mathbb{R}$. $\endgroup$– Fedor PetrovCommented Aug 29, 2016 at 8:25
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$\begingroup$ @FedorPetrov Thanks. Very interesting point. $\endgroup$– Ali TaghaviCommented Aug 29, 2016 at 11:00
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1 Answer
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No. Let $\langle A \rangle$ be a proper Borel subgroup of ${\mathbb R}$ generated by an algebraically independent Borel set $A$ with the cardinality of the continuum (such a set was constructed in
J. v. Neumann, MR 1512442 Ein System algebraisch unabhängiger zahlen, Math. Ann. 99 (1928), no. 1, 134--141.
, see also the discussion in
Barthélemy Le Gac, MR 687640 Some properties of Borel subgroups of real numbers, Proc. Amer. Math. Soc. 87 (1983), no. 4, 677--680.
). By the Steinhaus lemma, $\langle A \rangle$ has measure zero, thus $\langle A' \rangle$ is a Lebesgue measurable subgroup for every $A' \subset A$. But this gives $2^{\mathfrak c}$ such subgroups, more than the cardinality of the Borel $\sigma$-algebra, so at least one of these subgroups must be non-Borel.
answered Aug 29, 2016 at 8:52
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$\begingroup$ Is it obvious that the $\mathbb{Q}$-span of a Borel set, is a Borel set? $\endgroup$ Commented Sep 2, 2016 at 10:55
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1$\begingroup$ That's probably not true in general, but the construction of von Neumann is sigma-compact, and this will generate another sigma-compact set (as discussed in the paper of Le Gac). $\endgroup$ Commented Sep 3, 2016 at 5:41